Find the second least positive integer \(m\) such that \(m^3\) is the sum of \(m\) squares of consecutive integers.

Details and Assumptions

\(m>1\)

The least value is \(m=47\) where the consecutive integers start from \(22\), i.e. \(47^3=(21+1)^2+(21+2)^2\ldots(21+47)^2\)

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